First, multiply each side of the equation by the appropriate form of #1# to eliminate the fractions while keeping the equation balanced:

#color(red)(16) xx -27 = color(red)(16)((3v - 6)/8 - (8v + 9)/2)#

#-432 = (color(red)(16) xx (3v - 6)/8) - (color(red)(16) xx (8v + 9)/2)#

#-432 = (cancel(color(red)(16))color(red)(2) xx (3v - 6)/color(red)(cancel(color(black)(8)))) - (cancel(color(red)(16))color(red)(8) xx (8v + 9)/color(red)(cancel(color(black)(2))))#

#-432 = color(red)(2)(3v - 6) - color(red)(8)(8v + 9)#

#-432 = (color(red)(2) xx 3v) - (color(red)(2) xx 6) - (color(red)(8) xx 8v) + (color(red)(8) xx 9)#

#-432 = 6v - 12 - 64v - 72#

Next, group and combine like terms on the right side of the equation:

#-432 = 6v - 64v - 12 - 72#

#-432 = (6 - 64)v + (-12 - 72)#

#-432 = -58v + (-84)#

#-432 = -58v - 84#

Then, add #color(red)(84)# to each side of the equation to isolate the #v# term while keeping the equation balanced:

#-432 + color(red)(84) = -58v - 84 + color(red)(84)#

#-348 = -58v - 0#

#-348 = -58v#

Now, divide each side of the equation by #color(red)(-58)# to solve for #v# while keeping the equation balanced:

#(-348)/color(red)(-58) = (-58v)/color(red)(-58)#

#6 = (color(red)(cancel(color(black)(-58)))v)/cancel(color(red)(-58))#

#6 = v#

#v = 6#